12.5: Counting
Counting? You already know how to count or you wouldn't be taking a college-level math class, right? Well yes, but what we'll really be investigating here are ways of counting efficiently . When we get to the probability situations a bit later in this chapter we will need to count some very large numbers, like the number of possible winning lottery tickets. One way to do this would be to write down every possible set of numbers that might show up on a lottery ticket, but believe me: you don't want to do this.
Basic Counting
We will start, however, with some more reasonable sorts of counting problems in order to develop the ideas that we will soon need.
Suppose at a particular restaurant you have three choices for an appetizer (soup, salad or breadsticks) and five choices for a main course (hamburger, sandwich, quiche, fajita or pizza). If you are allowed to choose exactly one item from each category for your meal, how many different meal options do you have?
Solution
Solution 1 : One way to solve this problem would be to systematically list each possible meal:
\(\begin{array}{lll} \text{soup + hamburger} & \text{soup + sandwich} & \text{soup + quiche} \\ \text{soup + fajita} & \text{soup + pizza} & \text{salad + hamburger} \\ \text{salad + sandwich} & \text{salad + quiche} & \text{salad + fajita} \\ \text{salad + pizza} & \text{breadsticks + hamburger} & \text{breadsticks + sandwich} \\ \text{breadsticks + quiche} & \text{breadsticks + fajita} & \text{breadsticks + pizza} \end{array}\)
Assuming that we did this systematically and that we neither missed any possibilities nor listed any possibility more than once, the answer would be 15. Thus you could go to the restaurant 15 nights in a row and have a different meal each night.
Solution 2 : Another way to solve this problem would be to list all the possibilities in a table:
\(\begin{array}{|l|l|l|l|l|l|}
\hline & \textbf { hamburger } & \textbf { sandwich } & \textbf { quiche } & \textbf { fajita } & \textbf { pizza } \\
\hline \textbf { soup } & \text { Soup + burger } & & & & \\
\hline \textbf { salad } & \text { Salad + burger } & & & & \\
\hline \textbf { bread } & e t c . & & & & \\
\hline
\end{array}\)
In each of the cells in the table we could list the corresponding meal: soup + hamburger in the upper left corner, salad + hamburger below it, etc. But if we didn't really care what the possible meals are, only how many possible meals there are, we could just count the number of cells and arrive at an answer of 15, which matches our answer from the first solution. (It's always good when you solve a problem two different ways and get the same answer!)
Solution 3 : We already have two perfectly good solutions. Why do we need a third? The first method was not very systematic, and we might easily have made an omission. The second method was better, but suppose that in addition to the appetizer and the main course we further complicated the problem by adding desserts to the menu: we've used the rows of the table for the appetizers and the columns for the main courses—where will the desserts go? We would need a third dimension, and since drawing 3-D tables on a 2-D page or computer screen isn't terribly easy, we need a better way in case we have three categories to choose form instead of just two.
So, back to the problem in the example. What else can we do? Let's draw a tree diagram :
This is called a "tree" diagram because at each stage we branch out, like the branches on a tree. In this case, we first drew five branches (one for each main course) and then for each of those branches we drew three more branches (one for each appetizer). We count the number of branches at the final level and get (surprise, surprise!) 15.
If we wanted, we could instead draw three branches at the first stage for the three appetizers and then five branches (one for each main course) branching out of each of those three branches.
OK, so now we know how to count possibilities using tables and tree diagrams. These methods will continue to be useful in certain cases, but imagine a game where you have two decks of cards (with 52 cards in each deck) and you select one card from each deck. Would you really want to draw a table or tree diagram to determine the number of outcomes of this game?
Let's go back to the previous example that involved selecting a meal from three appetizers and five main courses, and look at the second solution that used a table. Notice that one way to count the number of possible meals is simply to number each of the appropriate cells in the table, as we have done above. But another way to count the number of cells in the table would be multiply the number of rows (3) by the number of columns (5) to get 15. Notice that we could have arrived at the same result without making a table at all by simply multiplying the number of choices for the appetizer (3) by the number of choices for the main course (5). We generalize this technique as the basic counting rule :
If we are asked to choose one item from each of two separate categories where there are \(m\) items in the first category and \(n\) items in the second category, then the total number of available choices is \(m \cdot n\)
This is sometimes called the multiplication rule for probabilities.
There are 21 novels and 18 volumes of poetry on a reading list for a college English course. How many different ways can a student select one novel and one volume of poetry to read during the quarter?
Solution
There are 21 choices from the first category and 18 for the second, so there are \(21 \cdot 18 = 378\) possibilities.
The Basic Counting Rule can be extended when there are more than two categories by applying it repeatedly, as we see in the next example.
Suppose at a particular restaurant you have three choices for an appetizer (soup, salad or breadsticks), five choices for a main course (hamburger, sandwich, quiche, fajita or pasta) and two choices for dessert (pie or ice cream). If you are allowed to choose exactly one item from each category for your meal, how many different meal options do you have?
Solution
There are 3 choices for an appetizer, 5 for the main course and 2 for dessert, so there are \(3 \cdot 5 \cdot 2=30\) possibilities.
A quiz consists of 3 true-or-false questions. In how many ways can a student answer the quiz?
Solution
There are 3 questions. Each question has 2 possible answers (true or false), so the quiz may be answered in \(2 \cdot 2 \cdot 2=8\) different ways. Recall that another way to write \(2 \cdot 2 \cdot 2\) is \(2^{3}\) which is much more compact.
Suppose at a particular restaurant you have eight choices for an appetizer, eleven choices for a main course and five choices for dessert. If you are allowed to choose exactly one item from each category for your meal, how many different meal options do you have?
- Answer
-
\(8 \cdot 11 \cdot 5=440\) menu combinations
Permutations
In this section we will develop an even faster way to solve some of the problems we have already learned to solve by other means. Let's start with a couple examples.
How many different ways can the letters of the word MATH be rearranged to form a four-letter code word?
Solution
This problem is a bit different. Instead of choosing one item from each of several different categories, we are repeatedly choosing items from the same category (the category is: the letters of the word MATH) and each time we choose an item we do not replace it, so there is one fewer choice at the next stage: we have 4 choices for the first letter (say we choose A), then 3 choices for the second (M, T and H; say we choose H), then 2 choices for the next letter (M and T; say we choose M) and only one choice at the last stage (T). Thus there are \(4 \cdot 3 \cdot 2 \cdot 1=24\) ways to spell a code worth with the letters MATH.
In this example, we needed to calculate \(n \cdot(n-1) \cdot(n-2) \cdots 3 \cdot 2 \cdot 1\). This calculation shows up often in mathematics, and is called the factorial , and is notated \(n\)!
\(n !=n \cdot(n-1) \cdot(n-2) \cdots 3 \cdot 2 \cdot 1\)
How many ways can five different door prizes be distributed among five people?
Solution
There are 5 choices of prize for the first person, 4 choices for the second, and so on. The number of ways the prizes can be distributed will be \(5 !=5 \cdot 4 \cdot 3 \cdot 2 \cdot 1=120\) ways.
Now we will consider some slightly different examples.
A charity benefit is attended by 25 people and three gift certificates are given away as door prizes: one gift certificate is in the amount of $100, the second is worth $25 and the third is worth $10. Assuming that no person receives more than one prize, how many different ways can the three gift certificates be awarded?
Solution
Using the Basic Counting Rule, there are 25 choices for the person who receives the \(\$ 100\) certificate, 24 remaining choices for the \(\$ 25\) certificate and 23 choices for the \(\$ 10\) certificate, so there are \(25 \cdot 24 \cdot 23=13,800\) ways in which the prizes can be awarded.
Eight sprinters have made it to the Olympic finals in the 100-meter race. In how many different ways can the gold, silver and bronze medals be awarded?
Solution
Using the Basic Counting Rule, there are 8 choices for the gold medal winner, 7 remaining choices for the silver, and 6 for the bronze, so there are \(8 \cdot 7 \cdot 6=336\) ways the three medals can be awarded to the 8 runners.
Note that in these preceding examples, the gift certificates and the Olympic medals were awarded without replacement ; that is, once we have chosen a winner of the first door prize or the gold medal, they are not eligible for the other prizes. Thus, at each succeeding stage of the solution there is one fewer choice (25, then 24, then 23 in the first example; 8, then 7, then 6 in the second). Contrast this with the situation of a multiple choice test, where there might be five possible answers — A, B, C, D or E — for each question on the test.
Note also that the order of selection was important in each example: for the three door prizes, being chosen first means that you receive substantially more money; in the Olympics example, coming in first means that you get the gold medal instead of the silver or bronze. In each case, if we had chosen the same three people in a different order there might have been a different person who received the $100 prize, or a different goldmedalist. (Contrast this with the situation where we might draw three names out of a hat to each receive a $10 gift certificate; in this case the order of selection is not important since each of the three people receive the same prize. Situations where the order is not important will be discussed in the next section.)
We can generalize the situation in the two examples above to any problem without replacement where the order of selection is important . If we are arranging in order \(r\) items out of \(n\) possibilities (instead of 3 out of 25 or 3 out of 8 as in the previous examples), the number of possible arrangements will be given by
\(n \cdot(n-1) \cdot(n-2) \cdots(n-r+1)\)
If you don't see why \((n-r+1)\) is the right number to use for the last factor, just think back to the first example in this section, where we calculated \(25 \cdot 24 \cdot 23\) to get \(13,800 .\) In this case \(n=25\) and \(r=3,\) so \(n-r+1=25-3+1=23,\) which is exactly the right number for the final factor.
Now, why would we want to use this complicated formula when it's actually easier to use the Basic Counting Rule, as we did in the first two examples? Well, we won't actually use this formula all that often, we only developed it so that we could attach a special notation and a special definition to this situation where we are choosing \(r\) items out of \(n\) possibilities without replacement and where the order of selection is important . In this situation we write:
\(_{n} P_{r}=n \cdot(n-1) \cdot(n-2) \cdots(n-r+1)\)
We say that there are \(_{n} P_{r}\) permutations of size \(r\) that may be selected from among \(n\) choices without replacement when order matters .
It turns out that we can express this result more simply using factorials.
\(_{n} P_{r}=\frac{n !}{(n-r) !}\)
In practicality, we usually use technology rather than factorials or repeated multiplication to compute permutations.
I have nine paintings and have room to display only four of them at a time on my wall. How many different ways could I do this?
Solution
Since we are choosing 4 paintings out of 9 without replacement where the order of selection is important there are\(_9 P_{4}=9 \cdot 8 \cdot 7 \cdot 6=3,024\) permutations.
How many ways can a four-person executive committee (president, vice-president, secretary, treasurer) be selected from a 16-member board of directors of a non-profit organization?
Solution
We want to choose 4 people out of 16 without replacement and where the order of selection is important. So the answer is \(_{16} P_{4}=16 \cdot 15 \cdot 14 \cdot 13=43,680\) .
How many 5 character passwords can be made using the letters A through Z
- if repeats are allowed
- if no repeats are allowed
- Answer
-
There are 26 characters.
- \(26^{5}=11,881,376\).
- \({26} \mathrm{P}_{5}=26 \cdot 25 \cdot 24 \cdot 23 \cdot 22=7,893,600\)
Combinations
In the previous section we considered the situation where we chose \(r\) items out of \(n\) possibilities without replacement and where the order of selection was important . We now consider a similar situation in which the order of selection is not important.
A charity benefit is attended by 25 people at which three $50 gift certificates are given away as door prizes. Assuming no person receives more than one prize, how many different ways can the gift certificates be awarded?
Solution
Using the Basic Counting Rule, there are 25 choices for the first person, 24 remaining choices for the second person and 23 for the third, so there are \(25 \cdot 24 \cdot 23=13,800\) ways to choose three people. Suppose for a moment that Abe is chosen first, Bea second and Cindy third; this is one of the 13,800 possible outcomes. Another way to award the prizes would be to choose Abe first, Cindy second and Bea third; this is another of the 13,800 possible outcomes. But either way Abe, Bea and Cindy each get $50, so it doesn't really matter the order in which we select them. In how many different orders can Abe, Bea and Cindy be selected? It turns out there are 6:
ABC ACB BAC BCA CAB CBA
How can we be sure that we have counted them all? We are really just choosing 3 people out of 3, so there are \(3 \cdot 2 \cdot 1=6\) ways to do this; we didn't really need to list them all, we can just use permutations!
So, out of the 13,800 ways to select 3 people out of 25, six of them involve Abe, Bea and Cindy. The same argument works for any other group of three people (say Abe, Bea and David or Frank, Gloria and Hildy) so each three-person group is counted six times . Thus the 13,800 figure is six times too big. The number of distinct three-person groups will be \(\frac{13,800}{6} = 2300\).
We can generalize the situation in this example above to any problem of choosing a collection of items without replacement where the order of selection is not important . If we are choosing \(r\) items out of \(n\) possibilities (instead of 3 out of 25 as in the previous examples), the number of possible choices will be given by , and we could use this formula for computation. However this situation arises so frequently that we attach a special notation and a special definition to this situation where we are choosing \(r\) items out of \(n\) possibilities without replacement where the order of selection is not important .
\(_{n} C_{r}=\frac{_n P_{r}}{r P_{r}}\)
We say that there are \(_{n} C_{r}\) combinations of size \(r\) that may be selected from among \(n\) choices without replacement where order doesn’t matter .
We can also write the combinations formula in terms of factorials:
\(_{n} C_{r}=\frac{n !}{(n-r) ! r !}\)
A group of four students is to be chosen from a 35-member class to represent the class on the student council. How many ways can this be done?
Solution
Since we are choosing 4 people out of 35 without replacement where the order of selection is not important there are = \(_{35} C_{4}=\frac{35 \cdot 34 \cdot 33 \cdot 32}{4 \cdot 3 \cdot 2 \cdot 1}=52,360\) combinations.
The United States Senate Appropriations Committee consists of 29 members; the Defense Subcommittee of the Appropriations Committee consists of 19 members. Disregarding party affiliation or any special seats on the Subcommittee, how many different 19-member subcommittees may be chosen from among the 29 Senators on the Appropriations Committee?
- Answer
-
Order does not matter. \(_{29} \mathrm{C}_{19}=20,030,010\) possible subcommittees
In the preceding Try it Now problem we assumed that the 19 members of the Defense Subcommittee were chosen without regard to party affiliation. In reality this would never happen: if Republicans are in the majority they would never let a majority of Democrats sit on (and thus control) any subcommittee. (The same of course would be true if the Democrats were in control.) So let's consider the problem again, in a slightly more complicated form:
The United States Senate Appropriations Committee consists of 29 members, 15 Republicans and 14 Democrats. The Defense Subcommittee consists of 19 members, 10 Republicans and 9 Democrats. How many different ways can the members of the Defense Subcommittee be chosen from among the 29 Senators on the Appropriations Committee?
Solution
In this case we need to choose 10 of the 15 Republicans and 9 of the 14 Democrats. There are \(_{15} C_{10}=3003\) ways to choose the 10 Republicans and \(_{14} C_{9}=2002\) ways to choose the 9 Democrats. But now what? How do we finish the problem?
Suppose we listed all of the possible 10-member Republican groups on 3003 slips of red paper and all of the possible 9-member Democratic groups on 2002 slips of blue paper. How many ways can we choose one red slip and one blue slip? This is a job for the Basic Counting Rule! We are simply making one choice from the first category and one choice from the second category, just like in the restaurant menu problems from earlier.
There must be \(3003 \cdot 2002=6,012,006\) possible ways of selecting the members of the Defense Subcommittee.
Probability using Permutations and Combinations
We can use permutations and combinations to help us answer more complex probability questions
A 4 digit PIN number is selected. What is the probability that there are no repeated digits?
Solution
There are \(10 \text { possible values for each digit of the PIN (namely: } 0,1,2,3,4,5,6,7,8,9),\) so there are \(10 \cdot 10 \cdot 10 \cdot 10=10^{4}=10000\) total possible PIN numbers.
To have no repeated digits, all four digits would have to be different, which is selecting without replacement. We could either compute 10 \cdot 9 \cdot 8 \cdot 7, or notice that this is the same as the permutation \(_{10} P_{4}=5040\).
The probability of no repeated digits is the number of 4 digit PIN numbers with no repeated digits divided by the total number of 4 digit PIN numbers. This probability is \(\frac{_{10} P_{4}}{10^{4}}=\frac{5040}{10000}=0.504\)
In a certain state's lottery, 48 balls numbered 1 through 48 are placed in a machine and six of them are drawn at random. If the six numbers drawn match the numbers that a player had chosen, the player wins $1,000,000. In this lottery, the order the numbers are drawn in doesn’t matter. Compute the probability that you win the million-dollar prize if you purchase a single lottery ticket.
Solution
In order to compute the probability, we need to count the total number of ways six numbers can be drawn, and the number of ways the six numbers on the player’s ticket could match the six numbers drawn from the machine. Since there is no stipulation that the numbers be in any particular order, the number of possible outcomes of the lottery drawing is \(_{48} C_{6}=12,271,512\). Of these possible outcomes, only one would match all six numbers on the player’s ticket, so the probability of winning the grand prize is:
\(\frac{_6 C_{6}}{_{48} C_{6}}=\frac{1}{12271512} \approx 0.0000000815\)
In the state lottery from the previous example, if five of the six numbers drawn match the numbers that a player has chosen, the player wins a second prize of $1,000. Compute the probability that you win the second prize if you purchase a single lottery ticket.
Solution
As above, the number of possible outcomes of the lottery drawing is \(_{48} C_{6}=12,271,512\). In order to win the second prize, five of the six numbers on the ticket must match five of the six winning numbers; in other words, we must have chosen five of the six winning numbers and one of the 42 losing numbers. The number of ways to choose 5 out of the 6 winning numbers is given by \(_{6} C_{5}=6\) and the number of ways to choose 1 out of the 42 losing numbers is given by \(_{42} C_{1}=42\). Thus the number of favorable outcomes is then given by the Basic Counting Rule: \(_{6} C_{5} \cdot_{42} C_{1}=6 \cdot 42=252\). So the probability of winning the second prize is.
\(\frac{\left(_{6} C_{5}\right)\left(_{42} C_{1}\right)}{_{48} C_{6}}=\frac{252}{12271512} \approx 0.0000205\)
A multiple-choice question on an economics quiz contains 10 questions with five possible answers each. Compute the probability of randomly guessing the answers and getting 9 questions correct.
- Answer
-
There are \(5^{10}=9,765,625\) different ways the exam can be answered. There are 10 possible locations for the one missed question, and in each of those locations there are 4 wrong answers, so there are 40 ways the test could be answered with one wrong answer.
\(\mathrm{P}(9 \text { answers correct })=\frac{40}{5^{10}} \approx 0.0000041\) chance
Compute the probability of randomly drawing five cards from a deck and getting exactly one Ace.
Solution
In many card games (such as poker) the order in which the cards are drawn is not important (since the player may rearrange the cards in his hand any way he chooses); in the problems that follow, we will assume that this is the case unless otherwise stated. Thus we use combinations to compute the possible number of 5-card hands, \(_{52} C_{5}\). This number will go in the denominator of our probability formula, since it is the number of possible outcomes.
For the numerator, we need the number of ways to draw one Ace and four other cards (none of them Aces) from the deck. Since there are four Aces and we want exactly one of them, there will be \(_4 C_{1}\) ways to select one Ace; since there are 48 non-Aces and we want 4 of them, there will be \(_{48} C_{4}\)\) ways to select the four non-Aces. Now we use the Basic Counting Rule to calculate that there will be \(_{4} C_{1} \cdot _{48} C_{4}\) ways to choose one ace and four non-Aces.
Putting this all together, we have
\(P(\text {one Ace} )=\frac{\left(_{4} C_{1}\right)\left(_{48} C_{4}\right)}{_{52} C_{5}}=\frac{778320}{2598960} \approx 0.299\)
Compute the probability of randomly drawing five cards from a deck and getting exactly two Aces.
Solution
The solution is similar to the previous example, except now we are choosing 2 Aces out of 4 and 3 non-Aces out of 48; the denominator remains the same:
\(P(\text {two Aces})=\frac{\left(_{4} C_{2}\right)\left(_{48} C_{3}\right)}{_{52} C_{5}}=\frac{103776}{2598960} \approx 0.0399\)
It is useful to note that these card problems are remarkably similar to the lottery problems discussed earlier.
Compute the probability of randomly drawing five cards from a deck of cards and getting three Aces and two Kings.
- Answer
-
\(P(\text {three Aces and two Kings})=\frac{\left(_{4} C_{3}\right)\left(_{4} C_{2}\right)}{_{52} C_{5}}=\frac{24}{2598960} \approx 0.0000092\)
Birthday Problem
Let's take a pause to consider a famous problem in probability theory:
Suppose you have a room full of 30 people. What is the probability that there is at least one shared birthday?
Take a guess at the answer to the above problem.  That seems to be the intuitive answer (\(\frac{30}{365}\), perhaps?). Let's see if we should listen to our intuition. Let's start with a simpler problem, however.
Suppose three people are in a room. What is the probability that there is at least one shared birthday among these three people?
Solution
There are a lot of ways there could be at least one shared birthday. Fortunately there is an easier way. We ask ourselves “What is the alternative to having at least one shared birthday?” In this case, the alternative is that there are no shared birthdays. In other words, the alternative to “at least one” is having none . In other words, since this is a complementary event,
\(P(\text {at least one})=1-P(\text {none})\)
We will start, then, by computing the probability that there is no shared birthday. Let's imagine that you are one of these three people. Your birthday can be anything without conflict, so there are 365 choices out of 365 for your birthday. What is the probability that the second person does not share your birthday? There are 365 days in the year (let's ignore leap years) and removing your birthday from contention, there are 364 choices that will guarantee that you do not share a birthday with this person, so the probability that the second person does not share your birthday is \(\frac{364}{365}\). Now we move to the third person. What is the probability that this third person does not have the same birthday as either you or the second person? There are 363 days that will not duplicate your birthday or the second person's, so the probability that the third person does not share a birthday with the first two is \(\frac{363}{365}\).
We want the second person not to share a birthday with you and the third person not to share a birthday with the first two people, so we use the multiplication rule:
\(P(\text {no shared birthday})=\frac{365}{365} \cdot \frac{364}{365} \cdot \frac{363}{365} \approx 0.9918\)
and then subtract from 1 to get
\(P(\text {shared birthday})=1-P(\text {no shared birthday})=1-0.9918=0.0082\).
This is a pretty small number, so maybe it makes sense that the answer to our original problem will be small. Let's make our group a bit bigger.
Suppose five people are in a room. What is the probability that there is at least one shared birthday among these five people?
Solution
Continuing the pattern of the previous example, the answer should be
\(P(\text {shared birthday})=1-\frac{365}{365} \cdot \frac{364}{365} \cdot \frac{363}{365} \cdot \frac{362}{365} \cdot \frac{361}{365} \approx 0.0271\)
Note that we could rewrite this more compactly as
\(P(\text {shared birthday})=1-\frac{_{365}P_5}{365^{5}} \approx 0.0271\)
which makes it a bit easier to type into a calculator or computer, and which suggests a nice formula as we continue to expand the population of our group.
Suppose 30 people are in a room. What is the probability that there is at least one shared birthday among these 30 people?
Solution
Here we can calculate
\(P(\text { shared birthday })=1-\frac{_{365}P_{30}}{365^{30}} \approx 0.706\)
which gives us the surprising result that when you are in a room with 30 people there is a 70% chance that there will be at least one shared birthday!
If you like to bet, and if you can convince 30 people to reveal their birthdays, you might be able to win some money by betting a friend that there will be at least two people with the same birthday in the room anytime you are in a room of 30 or more people. (Of course, you would need to make sure your friend hasn't studied probability!) You wouldn't be guaranteed to win, but you should win more than half the time.
This is one of many results in probability theory that is counterintuitive; that is, it goes against our gut instincts. If you still don't believe the math, you can carry out a simulation. Just so you won't have to go around rounding up groups of 30 people, someone has kindly developed a Java applet so that you can conduct a computer simulation. Go to this web page: http://statweb.stanford.edu/~susan/surprise/Birthday.html , and once the applet has loaded, select 30 birthdays and then keep clicking Start and Reset. If you keep track of the number of times that there is a repeated birthday, you should get a repeated birthday about 7 out of every 10 times you run the simulation.
Suppose 10 people are in a room. What is the probability that there is at least one shared birthday among these 10 people?
- Answer
-
\(P(\text {shared birthday})=1-\frac{_{365}P_{10}}{365^{10}} \approx 0.117\)


35.12: The Central Nervous System - Spinal Cord
- Describe the structure and function of the spinal cord
Connecting to the brainstem and extending down the body through the spinal column is the spinal cord: a thick bundle of nerve tissue that carries information about the body to the brain and from the brain to the body. The spinal cord is contained within the bones of the vertebral column, but is able to communicate signals to and from the body through its connections with spinal nerves (part of the peripheral nervous system).  Myelinated axons (the part of neurons that send signals) compose the “white matter,” while neuron and glial cell bodies (neuronal “support” cells) compose the “grey matter.” Grey matter is also composed of interneurons, which connect two neurons, each located in different parts of the body. Axons and cell bodies in the dorsal (facing the back of the animal) spinal cord convey mostly sensory information from the body to the brain. Axons and cell bodies in the ventral (facing the front of the animal) spinal cord primarily transmit signals controlling movement from the brain to the body.
The spinal cord also controls motor reflexes. These reflexes are quick, unconscious movements, such as automatically removing a hand from a hot object. Reflexes are so fast because they involve local synaptic connections. For example, the knee reflex that a doctor tests during a routine physical is controlled by a single synapse between a sensory neuron and a motor neuron. While a reflex may only require the involvement of one or two synapses, synapses with interneurons in the spinal column transmit information to the brain to convey what happened (the knee jerked, or the hand was hot).
In the United States, there around 10,000 spinal cord injuries each year. Because the spinal cord is the information superhighway connecting the brain with the body, damage to the spinal cord can lead to paralysis. The extent of the paralysis depends on the location of the injury along the spinal cord and whether the spinal cord was completely severed. For example, if the spinal cord is damaged at the level of the neck, it can cause paralysis from the neck down, whereas damage to the spinal column further down may limit paralysis to the legs. Spinal cord injuries are notoriously difficult to treat because spinal nerves do not regenerate, although ongoing research suggests that stem cell transplants may be able to act as a bridge to reconnect severed nerves. Researchers are also looking at ways to prevent the inflammation that worsens nerve damage after injury. One such treatment is to pump the body with cold saline to induce hypothermia. This cooling can prevent swelling and other processes that are thought to worsen spinal cord injuries.
Contributions and Attributions
- OpenStax College, Biology. October 17, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m44749/latest...ol11448/latest . License : CC BY: Attribution
- proprioception. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/proprioception . License : CC BY-SA: Attribution-ShareAlike
- somatosensation. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/somatosensation . License : CC BY-SA: Attribution-ShareAlike
- sulcus. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/sulcus . License : CC BY-SA: Attribution-ShareAlike
- gyrus. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/gyrus . License : CC BY-SA: Attribution-ShareAlike
- corpus callosum. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/corpus_callosum . License : CC BY-SA: Attribution-ShareAlike
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Key Points
- The spinal cord consists of a butterfly-shaped area of grey matter, containing neuronal and glial cell bodies, surrounded by white matter that contains the axons of the neurons.
- Neurons at the back of the spinal cord ( dorsal ) generally transmit information from the body to the brain, while neurons at the front of the spinal cord ( ventral ) primarily transmit information from the brain to the body.
- The spinal cord controls reflexes, which are incredibly fast reactions to stimuli; the speed at which they operate is due to the fact that they involve only a local connection between neurons and are not relayed through the brain.
- Spinal cord injuries often result in paralysis; they do not heal, as spinal nerves lack the ability to regenerate.
Key Terms
- grey matter : a collection of cell bodies and (usually) dendritic connections, in contrast to white matter
- synapse : the junction between the terminal of a neuron and either another neuron or a muscle or gland cell, over which nerve impulses pass
- axon : long slender projection of a nerve cell that conducts nerve impulses away from the cell body to other neurons, muscles, and organs
- white matter : a region of the central nervous system containing myelinated nerve fibers and no dendrites
- interneuron : a multipolar neuron that connects afferent and efferent neurons


13.17: Dark Matter
Dark Matter is matter believed to exist from its observed gravitational effects. We have never seen any light or energy from Dark Matter, yet we can “see” the effects of Dark Matter. Dark matter neither emits nor absorbs light, which is our primary means of observation. And it appears that this unseen Dark Matter is the vast majority of mass in the Universe. How do we know Dark Matter is there in the first place? We can weight a galaxy — look at the gravitational effects on objects in a galaxy. Does it match up with what we are seeing? We can also observe the Mass-to-Light ratio: the amount of mass versus the amount of light, which we observe.
It is important to understand Dark Matter if we want to ponder the fate of the Universe. Astronomers believe there are two possible scenarios: the Universe will continue to expand or the Universe will collapse, called the Big Crunch . The Big Crunch will occur if there is no or not enough Dark Matter.
The Milky Way’s Mass-to Light Ratio
We only see about 10% of the mass of the Milky Way; 90% of the mass is unseen, thus, the term Dark Matter . In the 1930s, Fritz Zwicky first proposed Dark Matter in the galaxies. At first it was not readily accepted, but as astronomers realized the need to find the source for the Milky Way’s (and other galaxies) missing matter, the concept of Dark Matter made sense. Galaxy mass physics shows Dark Matter exists. The challenge now is devising a method and instrumentation to physically observe Dark Matter.


6.8C: Measurements of Microbial Mass
- Recall ways of measuring microbial mass
Bacterial growth follows three phases: the lag phase, the log phase, and the stationary phase. The measurement of an exponential bacterial growth curve in a batch culture was traditionally a part of the training of all microbiologists; the basic means requires bacterial enumeration (cell counting) by direct and individual (microscopic, flow cytometry), direct and bulk (biomass), indirect and individual (colony counting), or indirect and bulk (most probable number, turbidity, nutrient uptake) methods. Models reconcile theory with the measurements.
METHODS OF MEASUREMENT
There are several methods for measuring cell mass, including the gravimeter method which uses ordinary balances to weigh a sample (dry weight/ml) after the water has been removed.
An indirect method for calculating cell mass is turbidimetry . Cell cultures are turbid: they absorb some of the light and let the rest of it pass through. The higher the cell concentration is, the higher the turbidity. Spectrophotometers are electrical appliances that can measure turbidity very accurately. The culture is placed in a translucent cuvette; the cuvette is placed in the machine and the turbidity measured immediately. Simple mathematical formulae help convert the detected turbidity to cell concentration. Using spectrophotometry for measuring the turbidity of cultures is known as turbidometry . Note the difference in spelling: turbidimetry and turbidometry are not the same word.
In spectrophotometry, cultures usually do not need to be diluted, although above a certain cell density the results lose reliability. Of all the electrical appliances used for counting cells, a spectrophotometer is the cheapest and its operation the fastest and most straightforward. This has made spectrophotometry the methods of choice for quick measurements of bacterial growth and related applications. There are spectrophotometers in which several cuvettes can be inserted at one time, reducing work time even more. Additionally, there are spectrophotometers that require extremely small volumes of culture, as little as 1 microliter. This, combined with the stochastic nature of liquid cultures, enables only an estimation of cell numbers.
An additional method for the measurement of microbial mass is the quantification of cells in a culture by plating the cells on a petri dish . If the cells are efficiently distributed on the plate, it can be generally assumed that each cell will give rise to a single colony. The colonies can then be counted, and based on the known volume of culture that was spread on the plate the cell concentration can be calculated.
As is with counting chambers, cultures usually need to be heavily diluted prior to plating; otherwise, instead of obtaining single colonies that can be counted, a so-called “lawn” will form, resulting in thousands of colonies lying over each other. Additionally, plating is the slowest method of all: most microorganisms need at least 12 hours to form visible colonies.
Key Points
- Calculating the dry weight of a sample enables one to calculate the cell count, but the sensitivity is limited to samples containing more than 10E8 bacteria per milliliter.
- Spectrophotometry is an indirect method for calculating cell concentrations by measuring the changes in turbidity.
- Bacteria can also be counted by using the plating method, which is based on the number of colonies formed in Petri dishes containing specific growth media.
Key Terms
- spectrophotometry : A spectrophotometer is commonly used for the measurement of transmittance or reflectance of solutions. However they can also be designed to measure the diffusivity on any of the listed light ranges that usually cover around 200nm – 2500nm using different controls and calibrations. [2] Within these ranges of light, calibrations are needed on the machine using standards that vary in type depending on the wavelength of the photometric determination. [3]
- flow cytometry : A technique used to sort and classify cells by using fluorescent markers on their surface.
- gravimeter : An instrument used to measure local variations in the gravitational field


30.3: Astrobiology
By the end of this section, you will be able to:
- Describe the chemical building blocks required for life
- Describe the molecular systems and processes driving the origin and evolution of life
- Describe the characteristics of a habitable environment
- Describe some of the extreme conditions on Earth, and explain how certain organisms have adapted to these conditions
Scientists today take a multidisciplinary approach to studying the origin, evolution, distribution, and ultimate fate of life in the universe; this field of study is known as astrobiology . You may also sometimes hear this field referred to as exobiology or bioastronomy. Astrobiology brings together astronomers, planetary scientists, chemists, geologists, and biologists (among others) to work on the same problems from their various perspectives.
Among the issues that astrobiologists explore are the conditions in which life arose on Earth and the reasons for the extraordinary adaptability of life on our planet. They are also involved in identifying habitable worlds beyond Earth and in trying to understand in practical terms how to look for life on those worlds. Let’s look at some of these issues in more detail.
The Building Blocks of Life
While no unambiguous evidence for life has yet been found anywhere beyond Earth, life’s chemical building blocks have been detected in a wide range of extraterrestrial environments. Meteorites (which you learned about in Cosmic Samples and the Origin of the Solar System) have been found to contain two kinds of substances whose chemical structures mark them as having an extraterrestrial origin—amino acids and sugars. Amino acids are organic compounds that are the molecular building blocks of proteins. Proteins are key biological molecules that provide the structure and function of the body’s tissues and organs and essentially carry out the “work” of the cell. When we examine the gas and dust around comets, we also find a number of organic molecules—compounds that on Earth are associated with the chemistry of life.
Expanding beyond our solar system, one of the most interesting results of modern radio astronomy has been the discovery of organic molecules in giant clouds of gas and dust between stars. More than 100 different molecules have been identified in these reservoirs of cosmic raw material, including formaldehyde, alcohol, and others we know as important stepping stones in the development of life on Earth. Using radio telescopes and radio spectrometers, astronomers can measure the abundances of various chemicals in these clouds. We find organic molecules most readily in regions where the interstellar dust is most abundant, and it turns out these are precisely the regions where star formation (and probably planet formation) happen most easily (Figure 30.4).
Clearly the early Earth itself produced some of the molecular building blocks of life. Since the early 1950s, scientists have tried to duplicate in their laboratories the chemical pathways that led to life on our planet. In a series of experiments known as the Miller-Urey experiments , pioneered by Stanley Miller and Harold Urey at the University of Chicago, biochemists have simulated conditions on early Earth and have been able to produce some of the fundamental building blocks of life, including those that form proteins and other large biological molecules known as nucleic acids (which we will discuss shortly).
Although these experiments produced encouraging results, there are some problems with them. The most interesting chemistry from a biological perspective takes place with hydrogen-rich or reducing gases, such as ammonia and methane. However, the early atmosphere of Earth was probably dominated by carbon dioxide (as Venus’ and Mars’ atmospheres still are today) and may not have contained an abundance of reducing gases comparable to that used in Miller-Urey type experiments. Hydrothermal vents—seafloor systems in which ocean water is superheated and circulated through crustal or mantle rocks before reemerging into the ocean—have also been suggested as potential contributors of organic compounds on the early Earth, and such sources would not require Earth to have an early reducing atmosphere.
Both earthly and extraterrestrial sources may have contributed to Earth’s early supply of organic molecules, although we have more direct evidence for the latter. It is even conceivable that life itself originated elsewhere and was seeded onto our planet—although this, of course, does not solve the problem of how that life originated to begin with.
Hydrothermal vents are beginning to seem more likely as early contributors to the organic compounds found on Earth. Read about hydrothermal vents and watch videos and slideshows on these and other deep-sea wonders at the Woods Hole Oceanographic Institution website.
Try an interactive simulation of hydrothermal-vent circulation at the Dive and Discover website.
The Origin and Early Evolution of Life
The carbon compounds that form the chemical basis of life may be common in the universe, but it is still a giant step from these building blocks to a living cell. Even the simplest molecules of the genes (the basic functional units that carry the genetic, or hereditary, material in a cell) contain millions of molecular units, each arranged in a precise sequence. Furthermore, even the most primitive life required two special capabilities: a means of extracting energy from its environment, and a means of encoding and replicating information in order to make faithful copies of itself. Biologists today can see ways that either of these capabilities might have formed in a natural environment, but we are still a long way from knowing how the two came together in the first life-forms.
We have no solid evidence for the pathway that led to the origin of life on our planet except for whatever early history may be retained in the biochemistry of modern life. Indeed, we have very little direct evidence of what Earth itself was like during its earliest history—our planet is so effective at resurfacing itself through plate tectonics (see the chapter on Earth as a Planet) that very few rocks remain from this early period. In the earlier chapter on Cratered Worlds, you learned that Earth was subjected to a heavy bombardment—a period of large impact events—some 3.8 to 4.1 billion years ago. Large impacts would have been energetic enough to heat-sterilize the surface layers of Earth, so that even if life had begun by this time, it might well have been wiped out.
When the large impacts ceased, the scene was set for a more peaceful environment on our planet. If the oceans of Earth contained accumulated organic material from any of the sources already mentioned, the ingredients were available to make living organisms. We do not understand in any detail the sequence of events that led from molecules to biology, but there is fossil evidence of microbial life in 3.5-billion-year-old rocks, and possible (debated) evidence for life as far back as 3.8 billion years.
Life as we know it employs two main molecular systems: the functional molecules known as proteins, which carry out the chemical work of the cell, and information-containing molecules of DNA (deoxyribonucleic acid) that store information about how to create the cell and its chemical and structural components. The origin of life is sometimes considered a “chicken and egg problem” because, in modern biology, neither of these systems works without the other. It is our proteins that assemble DNA strands in the precise order required to store information, but the proteins are created based on information stored in DNA. Which came first? Some origin of life researchers believe that prebiotic chemistry was based on molecules that could both store information and do the chemical work of the cell. It has been suggested that RNA (ribonucleic acid) , a molecule that aids in the flow of genetic information from DNA to proteins, might have served such a purpose. The idea of an early “RNA world” has become increasingly accepted, but a great deal remains to be understood about the origin of life.
Perhaps the most important innovation in the history of biology, apart from the origin of life itself, was the discovery of the process of photosynthesis , the complex sequence of chemical reactions through which some living things can use sunlight to manufacture products that store energy (such as carbohydrates), releasing oxygen as one by-product. Previously, life had to make do with sources of chemical energy available on Earth or delivered from space. But the abundant energy available in sunlight could support a larger and more productive biosphere, as well as some biochemical reactions not previously possible for life. One of these was the production of oxygen (as a waste product) from carbon dioxide, and the increase in atmospheric levels of oxygen about 2.4 billion years ago means that oxygen-producing photosynthesis must have emerged and become globally important by this time. In fact, it is likely that oxygen-producing photosynthesis emerged considerably earlier.
Some forms of chemical evidence contained in ancient rocks, such as the solid, layered rock formations known as stromatolites , are thought to be the fossils of oxygen-producing photosynthetic bacteria in rocks that are almost 3.5 billion years old (Figure 30.5). It is generally thought that a simpler form of photosynthesis that does not produce oxygen (and is still used by some bacteria today) probably preceded oxygen-producing photosynthesis, and there is strong fossil evidence that one or the other type of photosynthesis was functioning on Earth at least as far back as 3.4 billion years ago.
The free oxygen produced by photosynthesis began accumulating in our atmosphere about 2.4 billion years ago. The interaction of sunlight with oxygen can produce ozone (which has three atoms of oxygen per molecule, as compared to the two atoms per molecule in the oxygen we breathe), which accumulated in a layer high in Earth’s atmosphere. As it does on Earth today, this ozone provided protection from the Sun’s damaging ultraviolet radiation. This allowed life to colonize the landmasses of our planet instead of remaining only in the ocean.
The rise in oxygen levels was deadly to some microbes because, as a highly reactive chemical, it can irreversibly damage some of the biomolecules that early life had developed in the absence of oxygen. For other microbes, it was a boon: combining oxygen with organic matter or other reduced chemicals generates a lot of energy—you can see this when a log burns, for example—and many forms of life adopted this way of living. This new energy source made possible a great proliferation of organisms, which continued to evolve in an oxygen-rich environment.
The details of that evolution are properly the subject of biology courses, but the process of evolution by natural selection (survival of the fittest) provides a clear explanation for the development of Earth’s remarkable variety of life-forms. It does not, however, directly solve the mystery of life’s earliest beginnings. We hypothesize that life will arise whenever conditions are appropriate, but this hypothesis is just another form of the Copernican principle. We now have the potential to address this hypothesis with observations. If a second example of life is found in our solar system or a nearby star, it would imply that life emerges commonly enough that the universe is likely filled with biology. To make such observations, however, we must first decide where to focus our search.
Just how did life arise in the first place? And could it have happened with a different type of chemistry? Watch the 15-minute video Making Matter Come Alive in which a chemistry expert explores some answers to these questions, from a 2011 TED Talk.
Habitable Environments
Among the staggering number of objects in our solar system, Galaxy, and universe, some may have conditions suitable for life, while others do not. Understanding what conditions and features make a habitable environment —an environment capable of hosting life—is important both for understanding how widespread habitable environments may be in the universe and for focusing a search for life beyond Earth. Here, we discuss habitability from the perspective of the life we know. We will explore the basic requirements of life and, in the following section, consider the full range of environmental conditions on Earth where life is found. While we can’t entirely rule out the possibility that other life-forms might have biochemistry based on alternatives to carbon and liquid water, such life “as we don’t know it” is still completely speculative. In our discussion here, we are focusing on habitability for life that is chemically similar to that on Earth.
Life requires a solvent (a liquid in which chemicals can dissolve) that enables the construction of biomolecules and the interactions between them. For life as we know it, that solvent is water, which has a variety of properties that are critical to how our biochemistry works. Water is abundant in the universe, but life requires that water be in liquid form (rather than ice or gas) in order to properly fill its role in biochemistry. That is the case only within a certain range of temperatures and pressures—too high or too low in either variable, and water takes the form of a solid or a gas. Identifying environments where water is present within the appropriate range of temperature and pressure is thus an important first step in identifying habitable environments. Indeed, a “follow the water” strategy has been, and continues to be, a key driver in the exploration of planets both within and beyond our solar system.
Our biochemistry is based on molecules made of carbon, hydrogen, nitrogen, oxygen, phosphorus, and sulfur. Carbon is at the core of organic chemistry. Its ability to form four bonds, both with itself and with the other elements of life, allows for the formation of a vast number of potential molecules on which to base biochemistry. The remaining elements contribute structure and chemical reactivity to our biomolecules, and form the basis of many of the interactions among them. These “biogenic elements,” sometimes referred to with the acronym CHNOPS (carbon, hydrogen, nitrogen, oxygen, phosphorus, and sulfur), are the raw materials from which life is assembled, and an accessible supply of them is a second requirement of habitability.
As we learned in previous chapters on nuclear fusion and the life story of the stars, carbon, nitrogen, oxygen, phosphorus, and sulfur are all formed by fusion within stars and then distributed out into their galaxy as those stars die. But how they are distributed among the planets that form within a new star system, in what form, and how chemical, physical, and geological processes on those planets cycle the elements into structures that are accessible to biology, can have significant impacts on the distribution of life. In Earth’s oceans, for example, the abundance of phytoplankton (simple organisms that are the base of the ocean food chain) in surface waters can vary by a thousand-fold because the supply of nitrogen differs from place to place (Figure 30.6). Understanding what processes control the accessibility of elements at all scales is thus a critical part of identifying habitable environments.
With these first two requirements, we have the elemental raw materials of life and a solvent in which to assemble them into the complicated molecules that drive our biochemistry. But carrying out that assembly and maintaining the complicated biochemical machinery of life takes energy. You fulfill your own requirement for energy every time you eat food or take a breath, and you would not live for long if you failed to do either on a regular basis. Life on Earth makes use of two main types of energy: for you, these are the oxygen in the air you breathe and the organic molecules in your food. But life overall can use a much wider array of chemicals and, while all animals require oxygen, many bacteria do not. One of the earliest known life processes, which still operates in some modern microorganisms, combines hydrogen and carbon dioxide to make methane, releasing energy in the process. There are microorganisms that “breathe” metals that would be toxic to us, and even some that breathe in sulfur and breathe out sulfuric acid. Plants and photosynthetic microorganisms have also evolved mechanisms to use the energy in light directly.
Water in the liquid phase, the biogenic elements, and energy are the fundamental requirements for habitability. But are there additional environmental constraints? We consider this in the next section.
Life in Extreme Conditions
At a chemical level, life consists of many types of molecules that interact with one another to carry out the processes of life. In addition to water, elemental raw materials, and energy, life also needs an environment in which those complicated molecules are stable (don’t break down before they can do their jobs) and their interactions are possible. Your own biochemistry works properly only within a very narrow range of about 10 °C in body temperature and two-tenths of a unit in blood pH (pH is a numerical measure of acidity, or the amount of free hydrogen ions). Beyond those limits, you are in serious danger.
Life overall must also have limits to the conditions in which it can properly work but, as we will see, they are much broader than human limits. The resources that fuel life are distributed across a very wide range of conditions. For example, there is abundant chemical energy to be had in hot springs that are essentially boiling acid (see Figure 30.7). This provides ample incentive for evolution to fill as much of that range with life as is biochemically possible. An organism (usually a microbe) that tolerates or even thrives under conditions that most of the life around us would consider hostile, such as very high or low temperature or acidity, is known as an extremophile (where the suffix -phile means “lover of”). Let’s have a look at some of the conditions that can challenge life and the organisms that have managed to carve out a niche at the far reaches of possibility.
Both high and low temperatures can cause a problem for life. As a large organism, you are able to maintain an almost constant body temperature whether it is colder or warmer in the environment around you. But this is not possible at the tiny size of microorganisms; whatever the temperature in the outside world is also the temperature of the microbe, and its biochemistry must be able to function at that temperature. High temperatures are the enemy of complexity—increasing thermal energy tends to break apart big molecules into smaller and smaller bits, and life needs to stabilize the molecules with stronger bonds and special proteins. But this approach has its limits.
Nevertheless, as noted earlier, high-temperature environments like hot springs and hydrothermal vents often offer abundant sources of chemical energy and therefore drive the evolution of organisms that can tolerate high temperatures (see Figure 30.8); such an organism is called a thermophile . Currently, the high temperature record holder is a methane-producing microorganism that can grow at 122 °C, where the pressure also is so high that water still does not boil. That’s amazing when you think about it. We cook our food—meaning, we alter the chemistry and structure of its biomolecules—by boiling it at a temperature of 100 °C. In fact, food begins to cook at much lower temperatures than this. And yet, there are organisms whose biochemistry remains intact and operates just fine at temperatures 20 degrees higher.
Cold can also be a problem, in part because it slows down metabolism to very low levels, but also because it can cause physical changes in biomolecules. Cell membranes—the molecular envelopes that surround cells and allow their exchange of chemicals with the world outside—are basically made of fatlike molecules. And just as fat congeals when it cools, membranes crystallize, changing how they function in the exchange of materials in and out of the cell. Some cold-adapted cells (called psychrophiles ) have changed the chemical composition of their membranes in order to cope with this problem; but again, there are limits. Thus far, the coldest temperature at which any microbe has been shown to reproduce is about –25 ºC.
Conditions that are very acidic or alkaline can also be problematic for life because many of our important molecules, like proteins and DNA, are broken down under such conditions. For example, household drain cleaner, which does its job by breaking down the chemical structure of things like hair clogs, is a very alkaline solution. The most acid-tolerant organisms ( acidophiles ) are capable of living at pH values near zero—about ten million times more acidic than your blood (Figure 30.9). At the other extreme, some alkaliphiles can grow at pH levels of about 13, which is comparable to the pH of household bleach and almost a million times more alkaline than your blood.
High levels of salts in the environment can also cause a problem for life because the salt blocks some cellular functions. Humans recognized this centuries ago and began to salt-cure food to keep it from spoiling—meaning, to keep it from being colonized by microorganisms. Yet some microbes have evolved to grow in water that is saturated in sodium chloride (table salt)—about ten times as salty as seawater (Figure 30.10).
Very high pressures can literally squeeze life’s biomolecules, causing them to adopt more compact forms that do not work very well. But we still find life—not just microbial, but even animal life—at the bottoms of our ocean trenches, where pressures are more than 1000 times atmospheric pressure. Many other adaptions to environmental “extremes” are also known. There is even an organism, Deinococcus radiodurans, that can tolerate ionizing radiation (such as that released by radioactive elements) a thousand times more intense than you would be able to withstand. It is also very good at surviving extreme desiccation (drying out) and a variety of metals that would be toxic to humans.
From many such examples, we can conclude that life is capable of tolerating a wide range of environmental extremes—so much so that we have to work hard to identify places where life can’t exist. A few such places are known—for example, the waters of hydrothermal vents at over 300 °C appear too hot to support any life—and finding these places helps define the possibility for life elsewhere. The study of extremophiles over the last few decades has expanded our sense of the range of conditions life can survive and, in doing so, has made many scientists more optimistic about the possibility that life might exist beyond Earth.


1.1.0: Evolution of Atomic Theory
Learning Objectives
- Define: proton, neutron, electron, mass number, atomic number, isotope, and orbital.
- Differentiate between the three major subatomic particles by charge and location in an atom and determine how many of each an atom has.
- State the law of conservation of mass and describe how it relates to chemistry.
A Video Introduction to Atomic Theory through the Nineteenth Century From Crash Course Chemistry
Video \(\PageIndex{1}\): Lavoisier's discovery of The Law of Conservation of Matter led to the Laws of Definite and Multiple Proportions and eventually Dalton's Atomic Theory.
Atomic Theory through the Nineteenth Century
The earliest recorded discussion of the basic structure of matter comes from ancient Greek philosophers, the scientists of their day. In the fifth century BC, Leucippus and Democritus argued that all matter was composed of small, finite particles that they called atomos , a term derived from the Greek word for “indivisible.” They thought of atoms as moving particles that differed in shape and size, and which could join together. Later, Aristotle and others came to the conclusion that matter consisted of various combinations of the four “elements”—fire, earth, air, and water—and could be infinitely divided. Interestingly, these philosophers thought about atoms and “elements” as philosophical concepts, but apparently never considered performing experiments to test their ideas.
The Aristotelian view of the composition of matter held sway for over two thousand years, until English schoolteacher John Dalton helped to revolutionize chemistry with his hypothesis that the behavior of matter could be explained using an atomic theory. First published in 1807, many of Dalton’s hypotheses about the microscopic features of matter are still valid in modern atomic theory. Here are the postulates of Dalton’s atomic theory .
- Matter is composed of exceedingly small particles called atoms. An atom is the smallest unit of an element that can participate in a chemical change.
- An element consists of only one type of atom, which has a mass that is characteristic of the element and is the same for all atoms of that element (Figure \(\PageIndex{1}\)). A macroscopic sample of an element contains an incredibly large number of atoms, all of which have identical chemical properties.
- Atoms of one element differ in properties from atoms of all other elements.
- A compound consists of atoms of two or more elements combined in a small, whole-number ratio. In a given compound, the numbers of atoms of each of its elements are always present in the same ratio (Figure \(\PageIndex{2}\)).
- Atoms are neither created nor destroyed during a chemical change, but are instead rearranged to yield substances that are different from those present before the change (Figure \(\PageIndex{3}\)).
Figure \(\PageIndex{1}\): A pre-1982 copper penny (left) contains approximately 3 \(\times\) 10 22 copper atoms (several dozen are represented as brown spheres at the right), each of which has the same chemical properties. (credit: modification of work by “slgckgc”/Flickr)
Figure \(\PageIndex{2}\): Copper(II) oxide, a powdery, black compound, results from the combination of two types of atoms—copper (brown spheres) and oxygen (red spheres)—in a 1:1 ratio. (credit: modification of work by “Chemicalinterest”/Wikimedia Commons)
Figure \(\PageIndex{3}\) : When the elements copper (a shiny, red-brown solid, shown here as brown spheres) and oxygen (a clear and colorless gas, shown here as red spheres) react, their atoms rearrange to form a compound containing copper and oxygen (a powdery, black solid). (credit copper: modification of work by http://images-of-elements.com/copper.php ).
Dalton’s atomic theory provides a microscopic explanation of the many macroscopic properties of matter that you’ve learned about. For example, if an element such as copper consists of only one kind of atom, then it cannot be broken down into simpler substances, that is, into substances composed of fewer types of atoms. And if atoms are neither created nor destroyed during a chemical change, then the total mass of matter present when matter changes from one type to another will remain constant (the law of conservation of matter (or mass)).
Want to learn more about the Law of Conservation of Mass?
Video \(\PageIndex{2}\): "We are made of star stuff" - Carl Sagan .
Example \(\PageIndex{1}\): Test ing Dalton’s Atomic Theory
In the following drawing, the green spheres represent atoms of a certain element. The purple spheres represent atoms of another element. If the spheres touch, they are part of a single unit of a compound. Does the following chemical change represented by these symbols violate any of the ideas of Dalton’s atomic theory? If so, which one?
Solution
The starting materials consist of two green spheres and two purple spheres. The products consist of only one green sphere and one purple sphere. This violates Dalton’s postulate that atoms are neither created nor destroyed during a chemical change, but are merely redistributed. (In this case, atoms appear to have been destroyed.)
Exercise \(\PageIndex{1}\)
In the following drawing, the green spheres represent atoms of a certain element. The purple spheres represent atoms of another element. If the spheres touch, they are part of a single unit of a compound. Does the following chemical change represented by these symbols violate any of the ideas of Dalton’s atomic theory? If so, which one
- Answer
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The starting materials consist of four green spheres and two purple spheres. The products consist of four green spheres and two purple spheres. This does not violate any of Dalton’s postulates: Atoms are neither created nor destroyed, but are redistributed in small, whole-number ratios.
Dalton knew of the experiments of French chemist Joseph Proust, who demonstrated that all samples of a pure compound contain the same elements in the same proportion by mass . This statement is known as the law of definite proportions or the law of constant composition . The suggestion that the numbers of atoms of the elements in a given compound always exist in the same ratio is consistent with these observations. For example, when different samples of isooctane (a component of gasoline and one of the standards used in the octane rating system) are analyzed, they are found to have a carbon-to-hydrogen mass ratio of 5.33:1, as shown in Table \(\PageIndex{1}\).
| Sample | Carbon | Hydrogen | Mass Ratio |
|---|---|---|---|
| A | 14.82 g | 2.78 g | \(\mathrm{\dfrac{14.82\: g\: carbon}{2.78\: g\: hydrogen}=\dfrac{5.33\: g\: carbon}{1.00\: g\: hydrogen}}\) |
| B | 22.33 g | 4.19 g | \(\mathrm{\dfrac{22.33\: g\: carbon}{4.19\: g\: hydrogen}=\dfrac{5.33\: g\: carbon}{1.00\: g\: hydrogen}}\) |
| C | 19.40 g | 3.64 g | \(\mathrm{\dfrac{19.40\: g\: carbon}{3.63\: g\: hydrogen}=\dfrac{5.33\: g\: carbon}{1.00\: g\: hydrogen}}\) |
It is worth noting that although all samples of a particular compound have the same mass ratio, the converse is not true in general. That is, samples that have the same mass ratio are not necessarily the same substance. For example, there are many compounds other than isooctane that also have a carbon-to-hydrogen mass ratio of 5.33:1.00.
Dalton also used data from Proust, as well as results from his own experiments, to formulate another interesting law. The law of multiple proportions states that when two elements react to form more than one compound, a fixed mass of one element will react with masses of the other element in a ratio of small, whole numbers . For example, copper and chlorine can form a green, crystalline solid with a mass ratio of 0.558 g chlorine to 1 g copper, as well as a brown crystalline solid with a mass ratio of 1.116 g chlorine to 1 g copper. These ratios by themselves may not seem particularly interesting or informative; however, if we take a ratio of these ratios, we obtain a useful and possibly surprising result: a small, whole-number ratio.
\[\mathrm{\dfrac{\dfrac{1.116\: g\: Cl}{1\: g\: Cu}}{\dfrac{0.558\: g\: Cl}{1\: g\: Cu}}=\dfrac{2}{1}}\]
This 2-to-1 ratio means that the brown compound has twice the amount of chlorine per amount of copper as the green compound.
This can be explained by atomic theory if the copper-to-chlorine ratio in the brown compound is 1 copper atom to 2 chlorine atoms, and the ratio in the green compound is 1 copper atom to 1 chlorine atom. The ratio of chlorine atoms (and thus the ratio of their masses) is therefore 2 to 1 (Figure \(\PageIndex{4}\)).
Figure \(\PageIndex{4}\) : Compared to the copper chlorine compound in (a), where copper is represented by brown spheres and chlorine by green spheres, the copper chlorine compound in (b) has twice as many chlorine atoms per copper atom. (credit a: modification of work by “Benjah-bmm27”/Wikimedia Commons; credit b: modification of work by “Walkerma”/Wikimedia Commons)
Atomic Theory after the Nineteenth Century
If matter were composed of atoms, what were atoms composed of? Were they the smallest particles, or was there something smaller? In the late 1800s, a number of scientists interested in questions like these investigated the electrical discharges that could be produced in low-pressure gases, with the most significant discovery made by English physicist J. J. Thomson using a cathode ray tube. This apparatus consisted of a sealed glass tube from which almost all the air had been removed; the tube contained two metal electrodes. When high voltage was applied across the electrodes, a visible beam called a cathode ray appeared between them. This beam was deflected toward the positive charge and away from the negative charge, and was produced in the same way with identical properties when different metals were used for the electrodes. In similar experiments, the ray was simultaneously deflected by an applied magnetic field, and measurements of the extent of deflection and the magnetic field strength allowed Thomson to calculate the charge-to-mass ratio of the cathode ray particles. The results of these measurements indicated that these particles were much lighter than atoms (Figure \(\PageIndex{1}\)).
Figure \(\PageIndex{5}\) : (a) J. J. Thomson produced a visible beam in a cathode ray tube. (b) This is an early cathode ray tube, invented in 1897 by Ferdinand Braun. (c) In the cathode ray, the beam (shown in yellow) comes from the cathode and is accelerated past the anode toward a fluorescent scale at the end of the tube. Simultaneous deflections by applied electric and magnetic fields permitted Thomson to calculate the mass-to-charge ratio of the particles composing the cathode ray. (credit a: modification of work by Nobel Foundation; credit b: modification of work by Eugen Nesper; credit c: modification of work by “Kurzon”/Wikimedia Commons).
Based on his observations, here is what Thomson proposed and why: The particles are attracted by positive (+) charges and repelled by negative (−) charges, so they must be negatively charged (like charges repel and unlike charges attract); they are less massive than atoms and indistinguishable, regardless of the source material, so they must be fundamental, subatomic constituents of all atoms. Although controversial at the time, Thomson’s idea was gradually accepted, and his cathode ray particle is what we now call an electron , a negatively charged, subatomic particle with a mass more than one thousand-times less that of an atom. The term “electron” was coined in 1891 by Irish physicist George Stoney, from “ electr ic i on .”
In 1909, more information about the electron was uncovered by American physicist Robert A. Millikan via his “oil drop” experiments. Millikan created microscopic oil droplets, which could be electrically charged by friction as they formed or by using X-rays. These droplets initially fell due to gravity, but their downward progress could be slowed or even reversed by an electric field lower in the apparatus. By adjusting the electric field strength and making careful measurements and appropriate calculations, Millikan was able to determine the charge on individual drops (Figure \(\PageIndex{2}\)).
Figure \(\PageIndex{6}\): Millikan’s experiment measured the charge of individual oil drops. The tabulated data are examples of a few possible values.
Looking at the charge data that Millikan gathered, you may have recognized that the charge of an oil droplet is always a multiple of a specific charge, 1.6 \(\times\) 10 −19 C. Millikan concluded that this value must therefore be a fundamental charge—the charge of a single electron—with his measured charges due to an excess of one electron (1 times 1.6 \(\times\) 10 −19 C), two electrons (2 times 1.6 \(\times\) 10 −19 C), three electrons (3 times 1.6 \(\times\) 10 −19 C), and so on, on a given oil droplet. Since the charge of an electron was now known due to Millikan’s research, and the charge-to-mass ratio was already known due to Thomson’s research (1.759 \(\times\) 10 11 C/kg), it only required a simple calculation to determine the mass of the electron as well.
\[\mathrm{Mass\: of\: electron=1.602\times 10^{-19}\:\cancel{C}\times \dfrac{1\: kg}{1.759\times 10^{11}\:\cancel{C}}=9.107\times 10^{-31}\:kg} \tag{2.3.1}\]
Scientists had now established that the atom was not indivisible as Dalton had believed, and due to the work of Thomson, Millikan, and others, the charge and mass of the negative, subatomic particles—the electrons—were known. However, the positively charged part of an atom was not yet well understood. In 1904, Thomson proposed the “plum pudding” model of atoms, which described a positively charged mass with an equal amount of negative charge in the form of electrons embedded in it, since all atoms are electrically neutral. A competing model had been proposed in 1903 by Hantaro Nagaoka , who postulated a Saturn-like atom, consisting of a positively charged sphere surrounded by a halo of electrons (Figure \(\PageIndex{3}\)).
Figure \(\PageIndex{7}\) : (a) Thomson suggested that atoms resembled plum pudding, an English dessert consisting of moist cake with embedded raisins (“plums”). (b) Nagaoka proposed that atoms resembled the planet Saturn, with a ring of electrons surrounding a positive “planet.” (credit a: modification of work by “Man vyi”/Wikimedia Commons; credit b: modification of work by “NASA”/Wikimedia Commons).
The next major development in understanding the atom came from Ernest Rutherford , a physicist from New Zealand who largely spent his scientific career in Canada and England. He performed a series of experiments using a beam of high-speed, positively charged alpha particles (α particles) that were produced by the radioactive decay of radium; α particles consist of two protons and two neutrons (you will learn more about radioactive decay in the chapter on nuclear chemistry). Rutherford and his colleagues Hans Geiger (later famous for the Geiger counter) and Ernest Marsden aimed a beam of α particles, the source of which was embedded in a lead block to absorb most of the radiation, at a very thin piece of gold foil and examined the resultant scattering of the α particles using a luminescent screen that glowed briefly where hit by an α particle.
What did they discover? Most particles passed right through the foil without being deflected at all. However, some were diverted slightly, and a very small number were deflected almost straight back toward the source (Figure \(\PageIndex{4}\)). Rutherford described finding these results: “It was quite the most incredible event that has ever happened to me in my life. It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you” 1 (p. 68).
Figure \(\PageIndex{8}\): Geiger and Rutherford fired α particles at a piece of gold foil and detected where those particles went, as shown in this schematic diagram of their experiment. Most of the particles passed straight through the foil, but a few were deflected slightly and a very small number were significantly deflected.
Here is what Rutherford deduced: Because most of the fast-moving α particles passed through the gold atoms undeflected, they must have traveled through essentially empty space inside the atom. Alpha particles are positively charged, so deflections arose when they encountered another positive charge (like charges repel each other). Since like charges repel one another, the few positively charged α particles that changed paths abruptly must have hit, or closely approached, another body that also had a highly concentrated, positive charge. Since the deflections occurred a small fraction of the time, this charge only occupied a small amount of the space in the gold foil. Analyzing a series of such experiments in detail, Rutherford drew two conclusions:
- The volume occupied by an atom must consist of a large amount of empty space.
- A small, relatively heavy, positively charged body, the nucleus , must be at the center of each atom.
This analysis led Rutherford to propose a model in which an atom consists of a very small, positively charged nucleus, in which most of the mass of the atom is concentrated, surrounded by the negatively charged electrons, so that the atom is electrically neutral (Figure \(\PageIndex{5}\)).
Figure \(\PageIndex{9}\) : The α particles are deflected only when they collide with or pass close to the much heavier, positively charged gold nucleus. Because the nucleus is very small compared to the size of an atom, very few α particles are deflected. Most pass through the relatively large region occupied by electrons, which are too light to deflect the rapidly moving particles.
After many more experiments, Rutherford also discovered that the nuclei of other elements contain the hydrogen nucleus as a “building block,” and he named this more fundamental particle the proton , the positively charged, subatomic particle found in the nucleus. With one addition, which you will learn next, this nuclear model of the atom, proposed over a century ago, is still used today.
Another important finding was the discovery of isotopes. During the early 1900s, scientists identified several substances that appeared to be new elements, isolating them from radioactive ores. For example, a “new element” produced by the radioactive decay of thorium was initially given the name mesothorium. However, a more detailed analysis showed that mesothorium was chemically identical to radium (another decay product), despite having a different atomic mass. This result, along with similar findings for other elements, led the English chemist Frederick Soddy to realize that an element could have types of atoms with different masses that were chemically indistinguishable. These different types are called isotopes —atoms of the same element that differ in mass. Soddy was awarded the Nobel Prize in Chemistry in 1921 for this discovery.
One puzzle remained: The nucleus was known to contain almost all of the mass of an atom, with the number of protons only providing half, or less, of that mass. Different proposals were made to explain what constituted the remaining mass, including the existence of neutral particles in the nucleus. As you might expect, detecting uncharged particles is very challenging, and it was not until 1932 that James Chadwick found evidence of neutrons , uncharged, subatomic particles with a mass approximately the same as that of protons. The existence of the neutron also explained isotopes: They differ in mass because they have different numbers of neutrons, but they are chemically identical because they have the same number of protons. This will be explained in more detail later in this unit.
Video \(\PageIndex{2}\): An Introduction to Subatomic Particles
Atomic Structure
Figure \(\PageIndex{10}\) : Diagram of the structure of an atom including the nucleus with protons and neutrons and the outlying electrons.
Summary
Video \(\PageIndex{3}\): A summary of discoveries in atomic theory.
Video \(\PageIndex{4}\): A different summary of discoveries in atomic theory.
The ancient Greeks proposed that matter consists of extremely small particles called atoms. Dalton postulated that each element has a characteristic type of atom that differs in properties from atoms of all other elements, and that atoms of different elements can combine in fixed, small, whole-number ratios to form compounds. Samples of a particular compound all have the same elemental proportions by mass. When two elements form different compounds, a given mass of one element will combine with masses of the other element in a small, whole-number ratio. During any chemical change, atoms are neither created nor destroyed.
Although no one has actually seen the inside of an atom, experiments have demonstrated much about atomic structure. Thomson’s cathode ray tube showed that atoms contain small, negatively charged particles called electrons . Millikan discovered that there is a fundamental electric charge—the charge of an electron. Rutherford’s gold foil experiment showed that atoms have a small, dense, positively charged nucleus; the positively charged particles within the nucleus are called protons . Chadwick discovered that the nucleus also contains neutral particles called neutrons . Soddy demonstrated that atoms of the same element can differ in mass; these are called isotopes .
Footnotes
- Ernest Rutherford, “The Development of the Theory of Atomic Structure,” ed. J. A. Ratcliffe, in Background to Modern Science , eds. Joseph Needham and Walter Pagel, (Cambridge, UK: Cambridge University Press, 1938), 61–74. Accessed September 22, 2014, https://ia600508.us.archive.org/3/it...e032734mbp.pdf .
Glossary
- Dalton’s atomic theory
- set of postulates that established the fundamental properties of atoms
- law of constant composition
- (also, law of definite proportions) all samples of a pure compound contain the same elements in the same proportions by mass
- law of multiple proportions
- when two elements react to form more than one compound, a fixed mass of one element will react with masses of the other element in a ratio of small whole numbers
- law of definite proportions
- (also, law of constant composition) all samples of a pure compound contain the same elements in the same proportions by mass
- alpha particle (α particle)
- positively charged particle consisting of two protons and two neutrons
- electron
- negatively charged, subatomic particle of relatively low mass located outside the nucleus
- isotopes
- atoms that contain the same number of protons but different numbers of neutrons
- neutron
- uncharged, subatomic particle located in the nucleus
- proton
- positively charged, subatomic particle located in the nucleus
- nucleus
- massive, positively charged center of an atom made up of protons and neutrons
Contributors and Attributions
-
Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at<EMAIL_ADDRESS>).
- Adelaide Clark, Oregon Institute of Technology
- Crash Course Chemistry: Crash Course is a division of Complexly and videos are free to stream for educational purposes.
- TED-Ed’s commitment to creating lessons worth sharing is an extension of TED’s mission of spreading great ideas. Within TED-Ed’s growing library of TED-Ed animations, you will find carefully curated educational videos, many of which represent collaborations between talented educators and animators nominated through the TED-Ed website.


7.18: Body Language and Gestures
- Discuss strategies to effectively use body language and gestures to emphasize your message
One of the essential rules, and success factors, for public speaking is authenticity. This is as true for your non-verbal language as it is of the words you say and the ideas you express. Body language and gestures are a form of expression and can be either meaningful or distracting.
Toastmasters International, the global non-profit dedicated to teaching public speaking skills, believes that “gestures are probably the most evocative form of nonverbal communication a speaker can employ.” [1] In their Gestures: Your Body Speaks publication, they identify the following seven benefits of incorporating gestures into your speech: [2]
- Clarify and support your words
- Dramatize your ideas
- Lend emphasis and vitality to the spoken word
- Help dissipate nervous tension
- Function as visual aids
- Stimulate audience participation
- Are highly visible
Body language—how you dress as well as your mannerisms—is another powerful communication element. For perspective on this point, and a powerful speaking and life hack, watch social psychologist Amy Cuddy’s “ Your Body Language May Shape Who You Are ” TED Talk. The core idea is that we make judgments based on body language, and those judgments can predict meaningful life outcomes. In one example cited, social scientist Alex Todorov found that one-second judgments of political candidates’ faces predict 70 percent of U.S. Senate and gubernatorial race outcomes. What is perhaps more important, however, is that our body language reflects how we judge, think, and feel about ourselves. The key takeaway from this is that our bodies change our minds. That is, we can change not only how we are perceived but how we perceive ourselves by managing our body language. As a speaker, you must be conscious of, and cultivate, the presence you bring to your speech.
To quote Toastmasters International, “When you present a speech, you send two kinds of messages to your audience. While your voice is transmitting a verbal message, a vast amount of information is being visually conveyed by your appearance, your manner, and your physical behavior.” [3]
Your use of gestures and body movement should reflect not only your personal communication style but should also match the audience and the environment. A good practice is to “preview” the attendees or venue by sitting in on a prior event, watching a video, or scanning the event’s social feeds. This will give you a sense for audience dynamics and the size of the room. Certainly ask the event organizers in advance about the setup of the room in which you will speak. Consider adjusting your gestures to fit the audience, room size, and acoustics. For example, you may may want to tone down your gestures in a smaller space and put more emphasis on vocal rather than physical delivery. This doesn’t mean that you should put your personality on “mute” if you’re a naturally ebullient or expressive person. The key is to manage your mannerisms so they don’t overpower either your audience or your words. If the room is a large auditorium filled with enthusiastic fans, you may want to increase your physical presence with gestures to better “fill” the space. Rehearse new elements so they become fluid and reinforce rather than detract from your message. Remember that gestures and body language are most effective when they’re used as “visual punctuation.”
- Toastmasters International. Gestures: Your Body Speaks , p. 8 . 2011. Web. 26 Jun 2018.
- Ibid.
- Ibid.
Contributors and Attributions
- Body Language and Gestures. Authored by : Nina Burokas. Provided by : Lumen Learning. License : CC BY: Attribution
- Man gesturing while speaking. Authored by : Derrick Coetzee. Located at : https://commons.wikimedia.org/wiki/File:Frank_Schulenberg_speaking_and_gesturing_4.jpg . License : CC0: No Rights Reserved


2: Hardware
Learning Objectives
Upon successful completion of this chapter, you will be able to:
- Describe information systems hardware.
- Identify the primary components of a computer and the functions they perform.
- Explain the effect of the commoditization of the personal computer.
We discuss hardware and how it works. We will look at different types of computing devices, computer parts, learn how they interact and the effect of the commoditization of these devices.
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- 2.5: Other Computing Devices
- As personal computer technologies have become more commonplace, many of the components have been integrated into other devices that previously were purely mechanical. We have also seen an evolution in what defines a computer. Ever since the invention of the personal computer, users have clamored for a way to carry them around. Here we will examine several types of devices that represent the latest trends in personal computing.


3.6: Summary
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Key Takeaways
To describe motion in one dimension, we must define an axis with:
- An origin (where \(x=0\) ).
- A direction (the direction in which \(x\) increases).
- A unit for the length.
We describe the position of an object with a function \(x(t)\) that depends on time. The rate of change of position is called “velocity”, \(v_{x}(t)\) and is a function given by the time-derivative of position. The rate of change of velocity is called “acceleration”, \(a_{x}(t)\) and is a function given by the time-derivative of velocity.
Given a function for acceleration, \(a_x(t)\) , one can use its anti-derivative to determine velocity.
\(v_{x}(t)=\lim_{\Delta t\rightarrow 0} \frac{\Delta x}{\Delta t}=\frac{dx}{dt}\)
Given a function for velocity, \(v_x(t)\) , one can use its anti-derivative to determine position.
\(a_{x}(t)=\lim_{\Delta t\rightarrow 0}\frac{\Delta v}{\Delta t}=\frac{dv_{x}}{dt}\)
With a constant acceleration, \(a_{x}(t) = a_{x}\), if the object had velocity \(v_{0x}\) and position \(x_{0}\) at \(t = 0\):
\(\begin{aligned} v_{x}(t)&=v_{0x}t+a_{x}t \\[4pt] x(t)&=x_{0}+v_{0x}t+\frac{1}{2}a_{x}t^{2} \\[4pt] v^{2}-v_{0}^{2}&=2a(x-x_{0}) \end{aligned}\)
An inertial frame of reference is one that is moving with a constant velocity. It is impossible to define a frame of reference that is truly “at rest”, so we consider inertial frames of reference only relative to other frames of reference that we also consider to be inertial. If an object has a position \(x^A(t)\) in a given inertial frame of reference, \(x\) , that is moving with a velocity \(v^{'B}\) compared to a different inertial frame of reference, \(x^{'}\) , then the position of the object in the \(x^{'}\) frame of reference is given by \(x'(t)=v^{'B}+x^{A}(t)\) .
\(\begin{aligned} x^{'A}(t)&=v^{'B}t+x^{A}(t) \\[4pt] v^{'A}(t)&=v^{'B}+v^{A}(t) \\[4pt] a^{'A}(t)&=a(t) \end{aligned}\)
Important equations
If the position of an object is described by a function \(x(t)\) , then, its velocity, \(v_x(t)\) , and acceleration, \(a_x(t)\) , are given by:
\[\begin{aligned} v_x(t)&=\lim_{\Delta t\to 0}\frac{\Delta x}{\Delta t}=\frac{dx}{dt}\\[4pt] a_x(t)&=\lim_{\Delta t\to 0}\frac{\Delta v}{\Delta t}=\frac{dv_x}{dt}\\[4pt]\end{aligned}\]
Conversely, given the acceleration, \(a_x(t)\) , on can find the velocity and position:
\[\begin{aligned} v_x(t)=\lim_{\Delta t\to 0}\sum_ia_x(t_i)\Delta t+C=\int a_x(t)dt+C\\[4pt] x(t)=\lim_{\Delta t\to 0}\sum_iv_x(t_i)\Delta t+C=\int v_x(t)dt+C\\[4pt]\end{aligned}\]
With a constant acceleration, \(a_x(t)=a_x\) , if the object had velocity \(v_{0x}\) and position \(x_0\) at \(t=0\) :
\[\begin{aligned} v_x(t)&=v_{0x}t+a_xt\\[4pt] x(t)&=x_0+v_{0x}t+\frac{1}{2}a_xt^2 \\[4pt] v^{2}-v_{0}^{2}&=2a(x-x_{0}) \end{aligned}\]
If an object has position \(x^A\) as measured in a frame of reference \(x\) that is moving at constant speed \(v'^B\) as measured in a second frame of reference \(x'\) , then in the \(x'\) reference frame, the kinematic quantities for the object are obtained by the Galilean transformation:
\[\begin{aligned} x'^A(t) &= v'^Bt + x^A(t)\\[4pt] v'^A(t) &=v'^B+v^A(t)\\[4pt] a'^A(t) &= a(t)\end{aligned}\]
Important Definitions
Definition
Position : The distance between the defined coordinate system’s origin and an object. SI units: \([\text{m}]\). Common variable(s): \(\stackrel{\rightarrow}{x}, \stackrel{\rightarrow}{r}\).
Definition
Velocity : The rate at which position changes with respect to time. SI units: \([\text{ms}^{−1}]\). Common variable(s): \(\stackrel{\rightarrow}{v}\).
Definition
Acceleration : the rate at which velocity changes with respect to time. SI units: \([\text{ms}^{−2}]\). Common variable(s): \(\stackrel{\rightarrow}{a}\).